7.1 friction: basic applications - civil engineering friction: basic applications example 1, ......

7.1 Friction: Basic Applications

7.1 Friction: Basic Applications Example 1, page 1 of 2

Equations of equilibrium:

F x = 0: N

B f A = 0

F y = 0: N

A 98.1 N = 0

M A = 0: (98.1 N)(0.5 m) N

B(1.732 m) = 0

+

+

+

6

1. The uniform ladder is 2-m long and makes an angle of = 60

with the floor. If the wall at B is smooth and the coefficient of

static friction at A is A = 0.3, determine if the ladder can remain

in the position shown.

A

B

Mass of ladder = 10 kg

1 m

1 m

N B

N A

f A

(2 m)(sin 60°) = 1.732 m

Results of solving the above

equations of equilibrium:

N A = 98.1 N

N B = 28.3 N

f A = 28.3 N

7

Free-body diagramNo friction force is

present because the

wall is smooth.

Because the ladder is

uniform, the weight

acts through the center.

Weight = mg

= (10 kg)(9.81 m/s2 )

= 98.1 N

The friction force must

be drawn in a direction

opposing the

impending motion.

60°

12

3

5

4 Impending motion

A

B

(1 m)(cos 60°) = 0.5 m


Compute the maximum possible friction force that

the surfaces can develop at point A.

f A-max

AN A = (0.3)(98.1 N) = 29.4 N

To determine whether or not the ladder will stay in

the original position, the friction force found from the

equilibrium equations, f A = 28.3 N, must be

compared with the maximum force that the surfaces

at A can develop:

f A-max = 29.4 N

Since

f A = 28.3 N < 29.4 N = f

A-max

the surfaces are able to develop enough friction force

and the ladder will stay in equilibrium.

8



F x = 0: N

B f A = 0 (1)

F y = 0: N

A 98.1 N = 0 (2)

M A = 0: (98.1 N)(1 m) cos N

B(2 m) sin = 0 (3)

+

+

+

5

2. The uniform ladder is 2-m long and the wall at B

is smooth. If the coefficient of static friction at A is

A = 0.2, determine the smallest angle for which

the ladder can remain in the position shown.

A

B


1 m

1 m

N B

N A

f A

Three equations, but four unknowns: N B, f

A, N A, and

An additional equation is needed.

6

Free-body diagram

No friction force is

present because the

wall is smooth.

The friction force

must be drawn in a

direction opposing the

motion.

1

2

4

3 Impending motion

(1 m) cos

(2 m) sin

Weight = mg

= (10 kg)(9.81 m/s2 )

= 98.1 N


The fourth equation comes from the condition of

impending slip at point A, because if slip is just about to

occur, then the friction force, f A, is at its maximum value,

which is ANA:

f A f

A-max AN

A = (0.2)N A (4)

Three of the four equations are linear but the moment

equation, Eq. 3, is nonlinear because cos and sin

appear.

M A = 0: (98.1 N)(1m) cos

- N B(2 m) sin = 0 (Eq. 3 repeated)

The easiest way to solve these equations is to use the

general equation solver on a calculator. Alternatively,

manipulate the equations as follows.

First note that Eq. 2 implies that

N A = 98.1 N

Then using this value for N A in Eq. 4 gives

f A = N

A

= (0.2)(98.1 N)

= 19.62 N

+Using f

A = 19.62 N in Eq. 1 gives

N B = f

A

= 19.62 N

Using N B = 19.62 N in Eq. 3 gives

(98.1 N) cos (19.62 N)(2) sin = 0

Dividing through by cos and rearranging gives

Replacing the left-hand side of this equation by tan

gives

tan = 2.5

which implies

68.2 Ans.

8

sin 98.1cos (19.62)(2)

= = 2.5

7



F x = 0: N

B f A = 0 (1)

F y = 0: N

A + fB 98.1 N = 0 (2)

M A = 0: (98.1 N)(1 m) cos f

B(2 m) cos

N B(2 m) sin = 0 (3)

+

+


B

A

3. The uniform ladder is 2-m long. The

coefficient of static friction at A is A = 0.6 and

at B is B = 0.4. Determine the smallest angle, ,

for which the ladder can remain in the position

shown.

+

5

1 m

1 m

N B

N A

f A

Three equilibrium equations but five unknowns: N A,

N B, f

A, f B, and two more equations are needed.

6

Free-body diagram

Friction force, f B, is

present and opposes

the possible motion.1

3

4

(2 m) sin f B

Friction force

opposes the

possible motion.

(2 m) cos

(1 m) cos

Impending

motion

2

Weight = mg

= (10 kg)(9.81 m/s2 )

= 98.1 N


The two additional equations come from the condition of

impending slip at points A and B, because if slip is just about

to occur, then the friction forces, f A and f

B, are at their

maximum values, AN

A and N B

f A f

A-max AN

A = 0.6N A (4)

f B f

B-max BN

B = 0.4N B (5)

Four of the five equations are linear but the moment equation

Eq. 3 is nonlinear (sin and cos are present). To solve these

equations, use the general equation solver on your calculator or

manipulate the equation algebraically and use a trig identity

such as tan = sin /cos .

Results of solving the five equations (three equilibrium and

two friction equations) given above:

N A = 79.1 N

N B = 47.5 N

f A = 47.5 N

f B = 19.0 N

= 32.3° Ans.

7


4. Four round pegs A, B, C, and D are attached to

the bracket and loosely straddle the vertical pole.

When a 100-N force is applied as shown, the

bracket rotates slightly and friction forces

develop between pegs B, C, and the pole. If the

coefficient of static friction between the pegs and

the pole is determine the smallest value of

for which the bracket will support the load.

Neglect the effect of the rotation of the bracket

on the distances shown.

100 N

A

B

C

D

300 mm

50 mm

100 mm

Bracket rotates a

small amount

Peg A loses contact with the pole

Movement of bracket exaggerated for clarity1

100 N

D B

AC

Peg D loses contact with the pole


2

P = 100 N

100 mmf B

N B

N C

f C

Free-body diagram

300 mm 50 mm

Impending

motion of

bracket

3

The friction forces, f B and f

C,

resist the motion by pushing the

bracket up.

4

As the bracket inclines slightly, the pegs at A and D

lose contact with the pole. That is why no forces

appear at A and D on the free-body diagram.


F x = 0: N

B N C = 0 (1)

F y = 0: f

B + f C 100 N = 0 (2)

MC = 0: (100 N)(300 mm) + f B(50 mm)

N B(100 mm) = 0 (3)

++

+

7

6

The normal forces,

N B and N

C, are

directed from the

pole to the pegs .

5

B

C


There are only three equations of equilibrium but four

unknowns (f B, N

B, f C, and N

C), so at least one more

equation is needed. The additional equation comes from the

condition of impending slip at B, but if the bracket is going

to slip at B, it will also slip at C. So we have two additional

equations and one additional unknown, :

f B = f

B-max N B (4)

f C = f

C-max N C (5)

Solving Eqs. 1-5 gives the results below (Note that Eqs. 4

and 5 are nonlinear because multiplies N B and N

C):

f B = 50 N

N B = 325 N

f C = 50 N

N C = 325 N

= 0.154 Ans.

8


M

O y

O x

200 mm

N B

f B

Free-body diagram of

wheel.

1

Impending motion

of point on outer

surface of wheel

The friction force f B

opposes the motion.

2

3

Equation of moment equilibrium for the wheel

(Since we were not asked to compute the reactions

O x and O

y, we do not need to write the

force-equilibrium equations.):

M O = 0: f

B(200 mm) M = 0 (1)

4

+

O

200 mm

Radius = 200 mm 100 NC

300 mm

400 mm

B

A

M

O

5. Arm ABC acts as a brake on the wheel. The

coefficient of static friction at B is B = 0.4.

Determine the largest moment M that can act on the

wheel without causing it to slip.


The third equation follows from the condition

that slip impends at B:

f B f

B-max BN

B = 0.4N B (3)

Solving Eqs. 1-3 simultaneously yields

f B = 87.5 N

N B = 218.8 N

M = 17 500 N·mm = 17.5 N·m Ans.

A y

A x

A

5 Free-body diagram of arm ABC.

f B

N B

200 mm

400 mm

300 mm

100 N

7

+

Equation of moment equilibrium for the brake (Since we

were not asked to compute the reactions A x and A

y, we do

not need to write the force-equilibrium equations.):

M A = 0: f

B(200 mm) N B(400 mm)

+ (100 N)(300 mm + 400 mm) = 0 (2)

Thus far we have two equations but three unknowns (M, f B, and N

B), so another equation is needed.

8

6 The sense of the friction

force on the brake can be

determined by Newton's

Third Law (equal and

opposite to the force on

the wheel).


6. The uniform block is initially at rest when a 10-lb

force is applied. The coefficient of static friction

between the block and the plane is = 0.6.

Determine if the block will move.

1 ft

2 ft

10 lb

20 lb

A

(weight)


The resultant of the

distributed force fd is f.

5

The distributed friction force fd

opposes possible slip to the

right; the distributed normal

force Nd opposes possible

tipping of the block.

As the block is pushed to the right by the 10-lb force, the floor opposes the

possible motion by providing a distributed reaction force. The component of

this reaction force parallel to the floor is the distributed friction force fd, and

the component normal to the floor is the distributed normal force Nd.

3

2

The resultant of the distributed force

Nd is N. Because Nd opposes possible

tipping of the block, it is not uniform

but is greater near the right-hand side

of the base of the block to balance the

tendency to tip. Thus the resultant N

does not act at the middle of the base

but instead acts at some unknown

distance, x, from the middle.

6

B

Nd

fd

1 Free-body diagram

A

20 lb

10 lb

1 ft

0.5 ft

B

N

f

4 Free-body diagram showing resultant forces

2 ft

A

20 lb

10 lb

x



F x = 0: 10 lb f = 0

F y = 0: N 20 lb = 0

M A = 0: (20 lb)(0.5 ft) (10 lb)(2 ft) + N(0.5 ft + x) = 0

Solving these equations gives

f = 10 lb

N = 20 lb

x = 1 ft These are the values required if the system is to

stay in equilibrium, that is, not move. To

determine if the system can produce the 10-lb

friction force f required to keep the system in

equilibrium, we have to compare f with the

maximum possible value of the friction force:

f max N = (0.6)(20 lb) = 12 lb

Because f = 10 lb is less than the 12 lb maximum

possible force, the surfaces can develop enough

force to balance forces in the x direction (thus the

block will not slide to the right).

+

+

7

8

9

+


We next consider whether or not the block will tip.

Recall that solving the equilibrium equations gave the result

x = 1 ft. That is, to maintain equilibrium, the normal force

N must act at the location shown, 0.5 ft to the right of the

block. But this is impossible because N is the normal force

from the ground acting up on the block; the farthest N can

act is at the right hand corner, B. Thus the block will tip

because N cannot act far enough to the right to prevent it.

N (impossible location

because outside the base of

the block)0.5 ft 0.5 ft

x = 1 ft

20 lb

10

f

A B

10 lb


7. The uniform block is initially at rest when the force P

is applied. The coefficient of static friction between the

block and the plane is = 0.6. Determine the minimum

value of P that will cause the block to move.

2 ft

A

1 ft

(weight)20 lb

B

P


P1 ft

20 lb

A

Free-body diagram1

fd

N d

P

20 lb

A

2 ft

Free-body diagram showing resultant forces3

The resultant of the

distributed force fd is f.

f

4

As the block is pushed to the right by the force P, the floor opposes the

possible motion by providing a distributed reaction force. The component

of this reaction force parallel to the floor is the distributed friction force, fd,

and the component normal to the floor is the distributed normal force Nd.

2

The resultant of the distributed force Nd

is N. Because Nd opposes possible

tipping of the block, it is not uniform

but is greater near the right-hand side of

the base of the block to balance the

tendency to tip. Thus the resultant N

does not act at the middle of the base

but instead acts at some unknown

distance, x, from the middle.

5

N

B

B

0.5 ft x


1 ft

A B

20 lb

(weight)

P


F x = 0: P f = 0 (1)

F y = 0: N 20 lb = 0 (2)

M A = 0: (20 lb)(0.5 ft) P(2 ft)

+ N(0.5 ft + x) = 0 (3)

Three equations but four unknowns (P, f, N and x),

so one more equation is needed.

The fourth equation comes from considering

possible impending motion. There are two cases to

consider: sliding and tipping.

Case 1: Sliding

6

7

8

9+

++


P

20 lb

(weight)

BA

We have to analyze each case separately. Let's (arbitrarily)

choose Case 1 first. If sliding impends, then

f = f max N = 0.6N (4)

Solving Eqs. 1-4 simultaneously gives

P = 12 lb

N = 20 lb

f = 12 lb

x = 1.2 ft

Case 2: Tipping10 11

12


B

20 lb

P = 12 lb

A

0.5 ft

x = 1.2 ft

N = 20 lbf = 12 lb

But this diagram shows that the only way the equilibrium

equations for Case 1 can be satisfied is if the normal force N

lies to the right of the block (x = 1.2 ft). Since this is

impossible, the Case 1 assumption that sliding impends must

be incorrect.

Free body diagram for Case 1 (Sliding impends)13

14


f x

A

P

20 lb

B

Since the block is just about to tip, it loses contact with the

floor except at the corner B, where the normal force N is

concentrated. Since N acts at the corner, we know

x = 0.5 ft (5)

Solving the equilibrium equations, Eqs. 1, 2, and 3,

simultaneously with Eq. 5 gives

f = 5 lb

N = 20 lb

P = 5 lb Ans.

Since there were only two possibilities, sliding and tipping, and

we eliminated sliding, we know that the above result P = 5 lb is

correct. However, we can also check our work by verifying

that the friction force f is less than the maximum possible

value:

f = 5 lb < f max N = (0.6)(20 lb) = 12 lb. (OK)

Free body diagram for Case 2 (Tipping impends)15

160.5 ft 0.5 ft

N


8. The cylinder is initially at rest when a horizontal

force P is applied. The coefficients of static friction

at A and B are A = 0.3 and

B = 0.6. Determine

the minimum value of P that will cause the cylinder

to move.

0.3 m

P

20 kg

A

B

Radius = 0.2 m

A

0.3 m

P

B

f A

N A

f B

N B

Weight = mg

= (20 kg)(9.81 m/s2 )

= 196.2 N

Free-body diagram

Possible motion of

point B on cylinder.

Force P tends to

rotate the cylinder

clockwise.

The friction force from

the wall opposes the

motion of point B on

the cylinder.

Possible motion of point A.

The friction force from the floor

opposes the motion of point A on

the cylinder.

1

2

3

4

5

0.2 m

7.1 Friction: Basic Applications Example 8, page 2 of 3 +

6

Case 2Case 1

The cylinder rolls up the

wall without slipping.

The cylinder spins

about its center.

B

A

B

A

7 We have to analyze each case separately. Let's

(arbitrarily) choose Case 1 first. Thus if the cylinder

is about to slip about its center, then slip impends

simultaneously at points A and B, so

f A = f

A-max AN

A = 0.3N A (4)

f B = f

B-max BN

B = 0.6N B (5)


P = 554 N

f A = 34.6 N

f B = 312 N

N A = 115 N

N B = 519 N

A negative normal force, N A, is impossible (The

floor can't pull down on the cylinder), so the

assumption of slip at both A and B must be wrong.

Equilibrium equations

F x = 0: P + f

A N B = 0 (1)

F y = 0: 196.2 N + f

B + N A = 0 (2)

M A = 0: f

B(0.2 m) + NB(0.2 m) P(0.3 m) = 0 (3)

There are three equations and five unknowns (P, f A, N

A, f B, N

B),

so two more equations are needed. The two additional equations

come from considering possible impending motion. There are

two cases to consider:

+

+


Next consider Case 2 the cylinder is about to roll up the wall. Thus

the cylinder is about to lose contact with the floor at point A, and so

the friction and normal forces there are zero:

f A = 0 (6)

N A = 0 (7)

Solving the equilibrium equations, Eqs. 1, 2, and 3, simultaneously

with Eqs. 6 and 7 gives

f A = 0

N A = 0

f B = 196 N

N B = 392 N

P = 392 N Ans.

Since there were only two possibilities, spinning about the cylinder

center or rolling up the wall, and we eliminated spinning, the above

result P = 392 N must be correct. However, we can also check our

work by comparing the friction force, f B, with the maximum possible

value:

f B = 196 N < f

B-max BN

B = (0.6)(392 N) = 235 N (OK)

8


9. The small block B rests on top of the large block A. The

coefficients of static friction are shown in the figure.

Determine the smallest value of applied force P that will

keep block A from sliding down the inclined plane.

AB

60 kg

10 kg

P

Cord

Frictionless

pulley

B = 0.3

= 0.2

30°


Free-body diagram of block B

Tension in cord, T

Impending motion of

block B relative to

block A (If block A

moves down the plane,

block B must move up

the plane.)

Normal force

from block A

The friction force from block A opposes the

impending motion of block B up the incline.

The numerical value of

will be calculated later.

It's convenient to use an inclined

xy coordinate system.

P

B

30°

x

y

NB

fB

1

4

3

5

2

Weight = mg

= (10 kg)(9.81 m/s2)

= 98.1 N

+

+

Equations of equilibrium for block B. We assume that the blocks will

not tip because they are much longer than they are high; thus no

moment equation is needed (Since no dimensions are given, we could

not write a moment equation even if we wanted to).

F x =0: P T + f

B + (98.1 N) sin = 0 (1)

F y = 0: N

B (98.1 N) cos = 0 (2)

6


60°

30°

= 90° 60° = 30°

x

y

Geometry7

30°

NB

11

9

13

12

10

8

NA

fBT

y

x

= 30°

A

fA

Weight = mg

= (60 kg)(9.81 m/s2 )

= 588.6 N

[Weight of block A alone (Note

that the weight of block B is not

included because block B is not

part of this free-body. The effect

of the weight of block B is

transmitted through the normal

force, N B.)]

Impending motion

of block A relative

to block B.

Impending motion of block A

relative to inclined plane.

Normal force from

inclined plane

Friction force from

inclined plane opposes

motion of block A.

Friction force from block B opposes

motion of block A.

Free-body diagram of block A

30°


14 Equilibrium equations for block A

F x =0: (588.6 N) sin 30° fA f

B T = 0 (3)

F y = 0: (588.6 N) cos 30° + N

A N B = 0 (4)

Four equations in six unknowns (T, P, f A, N

A, f B, N

B). Two

more equations come from the condition of impending sliding

between the blocks and between block A and the plane:

f A f

A-max ANA = 0.2NA (5)

f B f

B-max BN

B = 0.3N B (6)


f A = 119 N

N A = 595 N

f B = 25 N

N B = 85 N

T = 150 N

P = 75 N Ans.

+

+

Free body diagram of block A repeated

fAA

x

yT

fB

NA

NB

30°

= 30°

588.6 N


10. The three blocks are stationary when the force P is applied. The

coefficients of static friction for each pair of surfaces are given in the

figure. Determine the smallest value of P for which motion will

occur. The blocks are sufficiently long that tipping will not occur.

Equilibrium equations for block A:

F x = 0: P f

AB = 0 (1)

F y = 0: N

AB 98.1 N = 0

The last equation gives

N AB = 98.1 N (2)

+

+

4

C

A

B

10 kg

10 kg

10 kg

AB = 0.8

BC = 0.3

C = 0.15

P

P A

N AB

f AB

Weight = mg

= (10 kg)(9.81 m/s2 )

= 98.1 N

Impending motion of

block A relative to B

Friction opposes the motion

2

3

Free-body diagram of block A1


Equilibrium equations for block B:

F x = 0: f

AB f BC = 0 (3)

F y = 0: 98.1 N 98.1 N + N

BC = 0


N BC = 196.2 N (4)

+

+

10

B

N AB = 98.1 N

f AB

Impending motion of block B

relative to block A (An observer

on A would see B moving in this

direction.)

Friction force opposes

relative motion

8 9

Free-body diagram of block B5

Weight = 98.1 N

f BC

N BC

6 Impending motion of

block B relative to C

7 Friction force opposes

relative motion


Equilibrium equations for block C:

F x = 0: f

BC f C = 0 (5)

F y = 0: 98.1 N 196.2 N + N

C = 0


N C = 294.3 N (6)

+

+

16

C

N BC = 196.2 N

f BC

Impending motion of block C

relative to block B (An observer

on B would see C moving in this

direction.)

Friction force opposes

relative motion

14

15

Free-body diagram of block C11

Weight = 98.1 N

f C

N C

12 Impending motion of

block C relative to

floor

13 Friction force opposes

relative motion


We now have six equilibrium equations but seven

unknowns (P, f AB, N

AB, f BC, N

BC, f C, N

C), so another

equation is needed.

17 The seventh equation comes from the condition of

impending slip. We have to consider three cases:18

A

B

C

Case 1

Impending

motion

Stationary

C Stationary

B

AImpending motion: blocks A and B move together

Case 2

C

B

A

Case 3

Impending

motion: blocks

A, B and C

move together


Slip impends so

f AB = f

AB-max

ABN

AB = (0.8)(98.1 N) = 78.5 N (7)

We have to check to see if the surfaces of contact between

blocks B and C develop enough friction force to keep block B

stationary.

20

Analyze each case separately.19

Stationary

C

A

B

Case 1

For block B, Eq. 3 is

f AB f

BC = 0

So for equilibrium,

f BC = f

AB

= 78.5 N

Let's compare this with the maximum possible friction

force:

f BC-max

BCN BC

= (0.3)(196.2 N) = 58.9 N (8)

So the surfaces can develop only 58.9 N while 78.5 N

are needed for equilibrium. Thus block B will move,

contrary to our assumption for Case 1.

by Eq. 2

21

22


Slip impends so

f BC = f

BC-max

= 58.9 N

We have to check to see if the surfaces of contact between

block C and the ground develop enough friction force to

keep block C stationary.

23For block C, Eq. 5 is

f BC f

C = 0

So for equilibrium,

f C = f

BC

= 58.9 N

Compare this with the maximum possible friction force

f C-max

CN C

= (0.15)(294.3 N) = 44.1 N (9)

So the surfaces can develop only 44.1 N while

58.9 N are needed for equilibrium. Thus block C will

move, contrary to our assumption for Case 2.

Case 2

C

B

A

Stationary

Impending

motion

together

by Eq. 8by Eq. 6

24

25


Slip impends so

f C = f

C-max = 44.1 N (10)

We don't have to check that the surfaces of contact

between blocks A and B and between B and C develop

enough friction to keep A and B in equilibrium, since there

were only three cases of possible motion, and we showed

that the first two cases were impossible. Nonetheless, we

can verify that our work is correct by showing that the

friction forces acting between A and B and between B and

C are less than their maximum possible values.

26 Eq. 5 gives f BC:

f BC f

C = 0

Thus

f BC = f

C

= 44.1 N (11)

and so

by Eq. 8

44.1 N = f BC < f

BC-max = 58.9 N (OK)

Eq. 3 gives f AB:

f AB f

BC = 0

Thus

f AB = f

BC

= 44.1 N (12)

and so

by Eq. 7

44.1 N = f AB < f

AB-max = 78.5 N (OK)

by Eq. 9

Case 3

B

C

AImpending

motion together

by Eq. 10

27

28


A

B

C

Thus the surfaces of contact between blocks A

and B and between B and C can develop

enough friction to keep blocks A, B, and C

moving together as a unit.

Finally, we can calculate P from Eq. 1:

P f AB = 0

or,

P = f AB = 44.1 N Ans.

Why didn't we consider a case like this?

Impending motion of A relative to B

Impending motion of B relative to C

Answer: No matter what the impending motion is,

there are only seven unknown forces (f AB, N

AB, f

BC, N

BC,

f C, N

C, and P). Since these seven unknowns must

satisfy the six equations of equilibrium, the unknowns

can be chosen to satisfy only one additional equation

a friction equation. In the unlikely event that the

masses and 's just happen to have values such that the

seven forces simultaneously satisfy the six equilibrium

equations and two friction equations, then one of the

eight equations must be redundant.

Applying this reasoning to Case 4, we see that if forces

exist that satisfy Case 4's equations, then these forces

must be identical to the forces satisfying the equations

for Case 1 (slip between A and B) and Case 2 (slip

between B and C). Since solving Case 4 would give

the same answer as solving Case 1 (or Case 2), we

don't have to consider Case 4. A similar argument can

be made for other possible motions. Case 4

Stationary

by Eq. 12

29

30

31

32


25°

11. The two cylinders shown are initially at rest when

horizontal forces of magnitude P/2 are applied to the ends

of the axle in the lower cylinder. The coefficients of

static friction for each pair of surfaces are given in the

figure. Determine the largest value of P that can be

applied without moving the cylinders up the inclined

plane.

A

B C

B = 0.4

C = 0.5

A = 0.6

P/2

Radius of each cylinder = 300 mm

Mass of each cylinder = 50 kg

P/2


y

x

NC

fC

C

O

fA

NA

25°

A

P

Free-body diagram of lower cylinder

The friction force from

the upper cylinder

opposes the relative

motion of point C on the

lower cylinder.

The friction force from the

plane opposes the motion

of point A on the cylinder.Impending motion of point A on

cylinder. The x component of

the applied force, P cos , is

pushing the cylinder up the

plane.

It is convenient to use an

inclined xy coordinate-system.

Radius = 300 mm

The numerical values of and will be

calculated later.

Weight = mg

= (50 kg)(9.81 m/s2 )

= 490.5 N

Impending motion of point C on lower

cylinder relative to upper cylinder (An

observer on the upper cylinder would

see this motion as the lower cylinder

moves).

1

6

8

7

2

4

3

5


Equilibrium equations for cylinder:

F x = 0: (490.5 N) sin P cos + f

A + N C = 0 (1)

F y = 0: 90.5 N) cos P sin + f

C + N A = 0 (2)

M O = 0: f

A(300 mm) f C(300 mm) = 0 (3)

Geometry

= 90° 65° = 25°

+

+

= 25°

y

x25°

65°

9

10

+

490.5 N

Radius = 300 mm

P

A

NA

fA

O

CfC

NC

x

y

25°

Free-body diagram of lower cylinder repeated


B

NB

fB

fC

C

P

= 25°

O

x

y Weight = 490.5 N

Free-body diagram of upper cylinder

Radius = 300 mm

The friction force

from the plane

opposes the motion

up the plane.Impending motion of point B as

normal force N C pushes the

upper cylinder up the plane.

Impending motion of point C on upper

cylinder relative to lower cylinder (An

observer on the lower cylinder would

see this motion as the upper cylinder

moves).

The friction force from the lower

cylinder opposes the relative

motion of point C on the upper

cylinder..

11

15

14

12

13

NC


Equilibrium equations

F x = 0: (490.5 N) sin 25° + f

B N C = 0 (4)

F y = 0: (490.5 N) cos 25° f

C + N B = 0 (5)

M O' = 0: f

B(300 mm) f C(300 mm) = 0 (6)

Thus far we have six equation but seven unknowns (P, f A, N

A,

f B, N

B, f C, N

C), so another equation is needed.

The seventh equation comes from the condition of impending

slip. We have to consider only two cases:

1. slip occurs at point B (and simultaneously rolling

occurs about points A and C).

2. slip occurs at point C (and simultaneously rolling

occurs about points A and B).

Slip at point A will be discussed later.

+

+

16

17

+

NC

Radius = 300 mm

490.5 N

y

x

O

25°

P

C

fC

fB

NB

B

Free-body diagram of upper cylinder repeated


Case 1

O'

O

A

B C

A

B

CO

O'

Displacement of point O

(Point O moves up the plane)

Before motion

After motion

Rolling without slipping (The radial

line OA on the lower cylinder

rotates through the same angle, , as

the radial line O'C on the upper

cylinder.)

Slip

For impending slip at B,

f B = f

B-max BN

B = 0.4N B (7)

20

19

18



f A = 296 N N

A = 618 N

f B = 296 N N

B = 741 N

f C = 296 N N

C = 504 N

P = 1111 N

We must check that the surfaces at A and C can provide enough friction

force to prevent slip and allow rolling:

f A 296 f

A-max AN

A = (0.6)(618 N) = 371 N (OK)

f C-max

CN C = (0.5)(504 N) = 252 N (Not enough! We need

f C = 296 N for equilibrium.)

So the assumption of impending slip at B is wrong.

21

22


Case 2 (Slip at C, rolling at A

and B)

O'

O

A

B C

A

B

C

O

O'

Displacement

of point O

Before motion

After motion

Rolling without

slipping

Slip

For impending slip at C,

f C = f

C-max CN

C = 0.5N C (8)

2624

23

Rolling without

slipping

25


Solving the six equilibrium equations, Eqs. 1-6, plus Eq. 8

yields

f A = 207 N N

A = 624 N

f B = 207 N N

B = 652 N

f C = 207 N N

C = 415 N

P = 915 N Ans.

The above answers must be correct since we eliminated

the only other possible case where slip impends. But we

can check our results by verifying that the friction forces

at A and B are less than their maximum possible values.

f A = 207 N f

A-max AN

A = (0.6)(624 N)

= 374 N (OK)

f B = 207 N f

B-max BN

B = (0.4)(652 N)

= 261 N (OK)

What about slip occurring at point A only? Well if the

lower cylinder moves, then the upper cylinder must also

move. But the only way that the upper cylinder can move

is if either 1) it slips at point B, or 2) it slips at point C.

Thus the case of slip impending at point A alone is

impossible and does not have to be considered.

What about simultaneous slip at A and B? Answer: we

have already found values of the seven unknowns in the

problem that satisfy the six equilibrium equations and the

equation for slip at B. In the unlikely case that the seven

values happen to satisfy an eighth equation (slip at A), then

that equation must be redundant, and the solution for the

eight equations is the same as we have already found for

the seven equations.

An analogous statement can be made for the case of

simultaneous slip at A and C.

27 29

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7.1 friction: basic applications - civil engineering friction: basic applications example 1, ......

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